用C语言完成公式(am)的n次方的计算
C\u8bed\u8a00\u7f16\u7a0b\uff1a\u6309\u6240\u7ed9\u7684\u516c\u5f0f\u8ba1\u7b97\u7ec4\u5408\u6570\u5e76\u8f93\u51fa\u7ed3\u679c\uff1aCmn\uff08m\u5728\u4e0an\u5728\u4e0b\uff09=n!/[(n-m)!m!]
\u6211\u7406\u89e3\u662f\u8fd9\u6837\u7684\u8f93\u5165\u8f93\u51fa\uff0c\u5982\u679c\u662f\u8fd9\u6837\u7684\u8f93\u5165\u8f93\u51fa\u7684\u8bdd\uff0c\u4ee3\u7801\u5982\u4e0b\uff1a
#include
#define N 34
main()
{
int i,j,a[N][N],n;
for(i=0;i<N;i++)
a[i][0]=a[i][i]=1;
for(i=2;i<N;i++)
{
for(j=1;j<i;j++)
a[i][j]=a[i-1][j]+a[i-1][j-1];
}
for(i=0;i<N;i++)
while(scanf("%d",&n),n>=0)
{
for(i=0;i<n;i++)
printf("%d ",a[n][i]);
printf("%d\n",a[n][n]);
}
return 0;
}
\u601d\u8def\uff1a\u5b9a\u4e49\u4e00\u4e2a\u51fd\u6570fun(x,n)\u6c42x\u7684n\u6b21\u65b9\uff0c\u5373\u8fdb\u884cn\u6b21for\u5faa\u73af\u6267\u884cx\u7684\u7d2f\u6210\uff0c\u4e3b\u51fd\u6570\u8c03\u7528fun\u51fd\u6570\u3002
\u53c2\u8003\u4ee3\u7801\uff1a
#include int fun(int x,int n){int s=1;while(n--){s*=x;}return s;} int main(){int x=2,y=10;printf("%d\n",fun(2,10));return 0;}/*\u8fd0\u884c\u7ed3\u679c\uff1a\u6c422\u768410\u6b21\u65b9 1024*/
int main()
{
int a,m,n;
int am,i;
long sum;
printf("Please input a:\n");
scanf("%d",&a);
fflush(stdin);
printf("Please input m:\n");
scanf("%d",&m);
fflush(stdin);
printf("Please input n:\n");
scanf("%d",&n);
fflush(stdin);
if(a > 5 || m > 5 || n > 5)
{
printf("Input error\n");
return 0;
}
am = a*m;
sum = 1;
for( i = 0; i < n; i++)
{
sum = sum * am;
}
printf("answer = %ld\n",sum);
return 0;
}
看看符合要求不？
int function(int a,int m,int n){
int value=1, c;
c=a*m;
for(i=0;i<n;i++){
value=value*c;
}
return value;
}
调用即可
#include<stdio.h>
void main(){
int a,m,n,s,i,com=1;
printf("请依次输入不大于5的a,m,n,以空格或者回车结束:");
scanf("%d",&a);
scanf("%d",&m);
scanf("%d",&n);
s=a*m;
for(i=0;i<n;i++)
com*=s;
printf("结果为：%d\n",com);
}
这个有点简单啊，O(∩_∩)O~
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